Re: [xsl] Noobie: normalize <b><a>...</a></b> to <a><b>...</b></a>

> I'm converting a non-XML data-dump into XML,
If it's really non-XML data, then XSLT can't help you. I presume
(since you're asking here), that you've munged it into XML somehow.
> and the document contains examples of both
> <a><b>...</b></a>
> and
> <b><a>...</a></b>
> which (in this document) are equivalent. I'd like to use XSLT to
> convert all examples of the latter to the former, with the
> following caveats:
> 1. <b> can contain mixed text, in which case nothing should be changed.
> 2. <b><a>...</a></b> should be changed to <a><b>...</b></a> only if
> the <a>...</a> element is the unique child node of <b>...</b>
> How can I do this?
I am sure others can provide better methods, but I *think* the
following will do what you ask:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform&quot;
version="1.0">
<xsl:template match="/">
<xsl:apply-templates select="node()"/>
</xsl:template>
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="b[not(text())][a][not(*[2])]">
<a><b><xsl:apply-templates select="a" mode="slurp"/></b></a>
</xsl:template>
<xsl:template match="a" mode="slurp">
<xsl:apply-templates/>
</xsl:template>
</xsl:stylesheet>